Right, but it's probably good to make that point clear before tackling this question.
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Dylan MorelandJun 28 '12 at 21:02

Note. Even assuming that it all makes sense, $J/I\cong R/I$ does not imply $(R/I)/(J/I) = 0$! Take $R=\oplus_{i=1}^{\infty}\mathbb{Z}$, $J$ the sum of the even-indexed terms, and $I$ the sum of the multiple-of-4 index terms. Then $R/I\cong R$, $J/I\cong R$, but $(R/I)/(J/I)$ is isomorphic to $R$ again, not to $0$.
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Arturo MagidinJun 28 '12 at 21:25

2 Answers
2

You can only speak of $I/J$ if $J\subseteq I$, so your question of whether
$$(R/I)/(J/I) \cong (R/J)/(I/J)$$
only makes sense if both $I\subseteq J$ and $J\subseteq I$, or equivalently $I=J$. Of course, when that is the case, then the above statement is trivially true.

Let's use your specific example, where $R=\mathbb{Z}$, $I=3\mathbb{Z}$, and $J=5\mathbb{Z}$. There is no such thing as $I/J$, but you can talk about $$I/(I\cap J)=3\mathbb{Z}/15\mathbb{Z}=\{\bar{0},\bar{3},\bar{6},\bar{9},\bar{12}\}\cong \mathbb{Z}/5\mathbb{Z},$$
and
$$J/(I\cap J)=5\mathbb{Z}/15\mathbb{Z}=\{\bar{0},\bar{5},\bar{10}\}\cong \mathbb{Z}/3\mathbb{Z}.$$
However, note that (by the third isomorphism theorem)
$$(R/(I\cap J))/(I/(I\cap J))\cong R/I$$
$$(R/(I\cap J))/(J/(I\cap J))\cong R/J$$
and there is no reason for those two rings to be isomorphic in general.

As written, what you have does not make sense. $J/I$ requires $I\subseteq J$, which is not met in your example. If both $J/I$ and $I/J$ make sense as written, then this requires $I=J$, and so in both cases we have $$(R/I)/(J/I) = (R/I)/(I/I) \cong R/I =R/J \cong (R/J)/(J/J) = (R/J)/(I/J),$$
and we are fine, but I doubt this is what you meant.

Instead, we can consider $(R/I)/((J+I)/I)$ and $(R/J)/((I+J)/J)$, since $I+J$ is the smallest ideal that contains both $I$ and $J$.

If so, then the answer is "yes", and it follows from the Isomorphism Theorems. In both cases we have that the quotient is isomorphic to $R/(I+J)$.

(However, as noted by Zev and as I failed to notice, this would not agree with your claim that "$5\mathbb{Z}/3\mathbb{Z} = \{\overline{0}, \overline{5},\overline{10}\}$", since my interpretation would be to set it equal to $\mathbb{Z}/3\mathbb{Z}$; while it also has $3$ elements, since $(I+J)/I \cong J/(I\cap J)$, it is not the "same" quotient.)

@ClarkKent: The definition of quotient only applies then you have an ideal. An ideal must, among other things, be a subset. So you cannot talk about $J/I$ unless $I$ is an ideal of $J$, which requires $I$ to be a subset of $J$. What you are doing, instead, is looking at $J/(I\cap J)$, which does make sense, since $I\cap J$ is an ideal of $J$. (Note that we are talking about the quotient structure, not the equivalence relation; the equivalence relation you give is induced by $I\cap J$)
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Arturo MagidinJun 28 '12 at 21:21